Propanoic acid has a boiling point of 141 °c, propanamide has a boiling point of 213 °c, and propanal has a boiling point of 48 °c. rationalize the differences in boiling point between these three compounds.

Explanation:

As it is given that boiling point of propanamide is very high. So, reason for this is that easy formation of hydrogen bonds which are strong enough that we have to provide large amount of heat to break it.

As in , the hydrogen atoms which are present are positive in nature. Due to this they are able to form hydrogen bonds with the neighboring oxygen atom.

Hence, these bonds are so strong that high heat needs to given to break them.

A propanoic acid contain carboxylic group as the functional group. So, this group is also able to form hydrogen bonding as it forms a hydrogen bond between an acid group and hydroxyl group of neighboring molecule.

Hence, it will also require high heat to break the bond due to which there will be increase in boiling point.

In propanal, there is presence of aldehyde functional group and three carbon atoms chain which will not form strong bonding with the hydrogen atom of CHO. Due to this there will exist weak Vander waal's force that is not at all strong enough.

As a result, less energy will be needed to break the bonds in propanal. Hence, it has very low boiling point.

Stoichiometry is the ratio of reactants to products in a balanced chemical reaction equation. 
the stoichiometry in this case of H₂ to O₂ is 2:1.
This means that 2 moles of H₂ reacts with 1 mol of O₂.
we have been asked to calculate the volume of H₂ gas.
Since both O₂ and H₂ are gases, at standard conditions its stated that molar volume of gases is 22.4 L
This means that at standard temperature and pressure, 1 mol of any gas occupies a volume of 22.4 L
first we need to calculate the number of O₂ moles reacted;
22.4 L of gas - 1 mol of O₂
1 L of gas - 1/22.4 mol/L 
then 20 L of O₂ - 1/22.4 mol/L * 20 L = 0.89 mol
stoichiometry of H₂ to O₂ is 2:1
the number of H₂ moles = 0.89*2 = 1.79 mol
1 mol occupies 22.4 L
Therefore 1.79 mol = 22.4 L/mol * 1.79 mol = 40 L
Therefore it can be seen that stoichiometry applies to volumes as well.
volume of H₂ : O₂ = 40 L : 20 L = 2:1
volume and moles both can be determined by stoichiometry.
Volume of H₂ reacted = 40 L